OCR MEI Further Pure Core 2024 June — Question 7 5 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2024
SessionJune
Marks5
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TopicIntegration with Partial Fractions
TypeImproper integrals with discontinuity
DifficultyChallenging +1.2 This is a straightforward improper integral question requiring recognition of a discontinuity at x=2 and application of a standard limit technique. The integration itself is routine (power rule after rewriting as (x-2)^{-1/3}), and the limit evaluation is mechanical. While it requires understanding of improper integrals and careful notation, it follows a standard template with no novel problem-solving requiredβ€”typical of Further Maths but not exceptionally challenging within that context.
Spec4.08c Improper integrals: infinite limits or discontinuous integrands
7
  1. Explain why \(\int _ { 1 } ^ { 2 } \frac { 1 } { \sqrt [ 3 ] { x - 2 } } \mathrm {~d} x\) is an improper integral.
  2. In this question you must show detailed reasoning. Use an appropriate limit argument to evaluate this integral.
Question 7:
AnswerMarks Guidance
7(a) 1
Because is not defined when x = 2
AnswerMarks Guidance
3 x βˆ’ 2B1
[1]2.4 isw. Must explicitly refer to π‘₯ = 2. See appendix.
DR
 1 3 2
 dx= (xβˆ’2)3
(xβˆ’2) 1 3 2
π‘Ž 1 3 2 π‘Ž
lim∫ dπ‘₯ or lim [ (π‘Žβˆ’2)3]
π‘Žβ†’2 1 3√π‘₯βˆ’2 π‘Žβ†’2 2 1
3 2
lim [ (π‘Žβˆ’2)3] = 0
π‘Žβ†’2 2
2 1 3 2
[∫ dπ‘₯ =]0βˆ’ (1βˆ’2)3
3√π‘₯βˆ’2 2
1
3
= βˆ’
AnswerMarks
2B1*
B1
B1
AnswerMarks
B1dep2.1
2.1
2.4
AnswerMarks
2.2aIntroducing an algebraic limit for β€˜2’ in original integral or
in integral of form π‘˜(π‘Žβˆ’2)2/3. Soi by next B1. Do not
allow π‘₯ as a limit.
Clear limit argument used for π‘˜(π‘Žβˆ’2)2/3 as π‘Ž β†’ 2. Do
not allow π‘₯ used for π‘Ž. = or β†’ must be used correctly.
Integral must have been explicitly evaluated for both
3
limits. Do not accept β€œβ†’ βˆ’ ”. Withhold if two limit
2
arguments considered.
AnswerMarks Guidance
Alternative methodB1* Correct substitution and integration
Let 𝑒 = π‘₯βˆ’2
1 3 2
∫ d𝑒 = 𝑒3
3βˆšπ‘’ 2
π‘Ž 1
lim∫ du
π‘Žβ†’0 3βˆšπ‘’
AnswerMarks Guidance
βˆ’1or 3 2 π‘Ž
lim [ 𝑒3]
AnswerMarks Guidance
π‘Žβ†’0 2 βˆ’1B1 Introducing an algebraic limit for β€˜0’ in original integral in
terms of 𝑒 or in integral of form π‘˜π‘’2/3. Soi by next B1.
Do not allow 𝑒 as a limit.
3 2
lim [ π‘Ž3] = 0
AnswerMarks Guidance
π‘Žβ†’0 2B1 Clear limit argument used for π‘˜(π‘Ž)2/3 as π‘Ž β†’ 0. Do not
allow π‘₯ used for π‘Ž. = or β†’ must be used correctly.
0 1 3 2
[∫ d𝑒 =]0βˆ’ (βˆ’1)3
3βˆšπ‘’ 2
βˆ’1
B1*
Correct substitution and integration
3
= βˆ’
AnswerMarks Guidance
2B1dep Integral must have been explicitly evaluated for both
3
limits. Do not accept β€œβ†’ βˆ’ ”. Withhold if two limit
2
arguments considered.
[4]
Question 7:
7 | (a) | 1
Because is not defined when x = 2
3 x βˆ’ 2 | B1
[1] | 2.4 | isw. Must explicitly refer to π‘₯ = 2. See appendix.
DR
 1 3 2
 dx= (xβˆ’2)3
(xβˆ’2) 1 3 2
π‘Ž 1 3 2 π‘Ž
lim∫ dπ‘₯ or lim [ (π‘Žβˆ’2)3]
π‘Žβ†’2 1 3√π‘₯βˆ’2 π‘Žβ†’2 2 1
3 2
lim [ (π‘Žβˆ’2)3] = 0
π‘Žβ†’2 2
2 1 3 2
[∫ dπ‘₯ =]0βˆ’ (1βˆ’2)3
3√π‘₯βˆ’2 2
1
3
= βˆ’
2 | B1*
B1
B1
B1dep | 2.1
2.1
2.4
2.2a | Introducing an algebraic limit for β€˜2’ in original integral or
in integral of form π‘˜(π‘Žβˆ’2)2/3. Soi by next B1. Do not
allow π‘₯ as a limit.
Clear limit argument used for π‘˜(π‘Žβˆ’2)2/3 as π‘Ž β†’ 2. Do
not allow π‘₯ used for π‘Ž. = or β†’ must be used correctly.
Integral must have been explicitly evaluated for both
3
limits. Do not accept β€œβ†’ βˆ’ ”. Withhold if two limit
2
arguments considered.
Alternative method | B1* | Correct substitution and integration
Let 𝑒 = π‘₯βˆ’2
1 3 2
∫ d𝑒 = 𝑒3
3βˆšπ‘’ 2
π‘Ž 1
lim∫ du
π‘Žβ†’0 3βˆšπ‘’
βˆ’1 | or | 3 2 π‘Ž
lim [ 𝑒3]
π‘Žβ†’0 2 βˆ’1 | B1 | Introducing an algebraic limit for β€˜0’ in original integral in
terms of 𝑒 or in integral of form π‘˜π‘’2/3. Soi by next B1.
Do not allow 𝑒 as a limit.
3 2
lim [ π‘Ž3] = 0
π‘Žβ†’0 2 | B1 | Clear limit argument used for π‘˜(π‘Ž)2/3 as π‘Ž β†’ 0. Do not
allow π‘₯ used for π‘Ž. = or β†’ must be used correctly.
0 1 3 2
[∫ d𝑒 =]0βˆ’ (βˆ’1)3
3βˆšπ‘’ 2
βˆ’1
B1*
Correct substitution and integration
3
= βˆ’
2 | B1dep | Integral must have been explicitly evaluated for both
3
limits. Do not accept β€œβ†’ βˆ’ ”. Withhold if two limit
2
arguments considered.
[4]
7
\begin{enumerate}[label=(\alph*)]
\item Explain why $\int _ { 1 } ^ { 2 } \frac { 1 } { \sqrt [ 3 ] { x - 2 } } \mathrm {~d} x$ is an improper integral.
\item In this question you must show detailed reasoning.

Use an appropriate limit argument to evaluate this integral.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2024 Q7 [5]}}
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